Integration by Partial Fractions

What would be the value of $\int \frac{x^2+1}{{\left(x-1\right)}^2\left(x+3\right)}dx$

Let $x\ne \frac{-3}{5},\frac{2}{5}$, if $f\left(\frac{2x+1}{5x+3}\right)=x+2$, then $\int f\left(x\right)dx=$

If $\int \frac{x+3}{{\left(x-1\right)}^2\left(2x-1\right)}dx=\frac{\mathrm{A}}{x-1}+B \log \left(2\mathrm{x}-1\right)+C \log \left(\mathrm{x}-1\right)+K$ then $A+B+C=$

$\int \frac{dx}{x^2-x}=$

$\int \frac{e^{5x}+e^x}{e^{6x}+1}dx=\frac{2}{3}f\left(x\right)+g\left(x\right)+c$, then $f\left(x\right)$ equals

The general solution of the differential equation $y^{\text{'}}=\frac{4}{x(x-4)}$ is 

$\int \frac{x^5+1}{x+1}dx=$_____$+C$

$\int \frac{x^3}{x+1}dx=$ 

$\int \frac{3x-1}{{\left(x+2\right)}^2}dx$ is equal to

The integral $\int \frac{32}{x^4-16}dx=$

$\int \frac{x-1}{(x-2)(x-3)}dx=$

If $\mathrm{p}, \mathrm{q}$ and $r$ are prime numbers such that ${\mathrm{p}}^2-{\mathrm{q}}^2=\mathrm{r},$ then $\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{px}+\mathrm{q}}$ is

The partial fractions of $\frac{3x^3-8x^2+10}{{\left(x-1\right)}^4}$ is

If $\int \frac{dx}{\left(x+2\right)\left(x^2+1\right)}=a\log \left(1+x^2\right)+b{\tan }^{-1}x+\frac{1}{5}\log \left(x+2\right)+C$ , then

Find: $\int \frac{xdx}{(x+1)(x+2)}$

$\int \frac{dx}{x^3+3x^2+2x}=$

The integral$\int \frac{(2\mathrm{sin\theta }-1)\mathrm{cos\theta }}{5-{\cos }^2\mathrm{\theta }-4\mathrm{sin\theta }}d\theta$  is equal to

(where $C$ is a constant of integration)